How many ways can a four digit number be formed with different restrictions?

[six789]

I just want to ask if my work is correct...
This are the questions:
1. In how many ways can the letters of MISSISSAUGA be arranged?
-->6*6*6*6*6*6*6*6*6*6*6 = 362,797,056

2. A man bought two vanilla ice creams, three chocolate cones, four strawberry cones and maple walnut cone for his 10 children. In how many ways can he distribute the flavours among the children?
-->4*4*4*4*4*4*4*4*4*4 = 1,048,576

3. a) how many permutations are there of the letters of the word BASKETBALL?
--> 7*7*7*7*7*7*7*7*7*7 = 282,475,249
b) how many of the arrangements begin with K?
--> 1*7*7*7*7*7*7*7*7*7 = 40,353,607
c) how may of the arrangements would the two L's be together?
--> 1*7*7*7*7*7*7*7*7 = 5,764,801

4. How many four digit numbers are there with the following retrictions?
a) using the digits 1 to 8
--> 8*7*7*7 = 2,744
b) using the digits 0 to 7
--> 7*6*6*6 = 1,512
c) using only odd digits
--> this i don't get, are you going to use 1 to 10? (i don't know)
d) the number is odd
--> same
e) the number is even
-->same

 

[Erzeon]

4a. shouldn't it be 8P4(dont know how to write it on the comp but it looks a bit like that on the calculator). therefore 8 x 7 x 6 x 5.

Bah my answers are wrong, I kept thinking about combinations and not permutations

I just looked at my old maths book and I just remembered haha.
1. is 11!/(4! x 2! x 2!)
2. 10!/(2! x 3! x 4! x 1!)
Just keep using the formula I wrote just down here.

The formula I used was, (no. of letters or whatever)!/((no. of same things eg Letter S)! x (no of other same things eg Letter A)! and keep multiplying any same samples) the book writes n!/(p!q!r!...)

 

[six789]

4a. shouldn't it be 8P4(dont know how to write it on the comp but it looks a bit like that on the calculator). therefore 8 x 7 x 6 x 5.

for this, i think you are right...

 

[six789]

The formula I used was, (no. of letters or whatever)!/((no. of same things eg Letter S)! x (no of other same things eg Letter A)! and keep multiplying any same samples) the book writes n!/(p!q!r!...)

the formula you gave to me is very helpful... coz our teacher doesn't write anything on the board, he just keeps on talking...lol...thanks

what if you got words with same letters with combination of different ones?

 

[Erzeon]

I added Question 2 answer and for 4c. It should be 5P4. that is 5!/1! since there are only 5 odd numbers from 0-9 (i.e. 1,3,5,7,9). 10 is not a digit, its a combination of 2 digits.

4d means that the last digit has to be odd.

 

[Erzeon]

what if you got words with same letters with combination of different ones?

can you please give me an example, i don't quite get what you mean.

 

[six789]

thanks erzeon for helping me this time!

 

[six789]

can you please give me an example, i don't quite get what you mean.

i get it now, since you said that just keeping multiplying the same letters... and since it is only 1 letter, it is only 1!, so it is understood that it is one... i get waht you mean...

 

[six789]

I added Question 2 answer and for 4c. It should be 5P4. that is 5!/1! since there are only 5 odd numbers from 0-9 (i.e. 1,3,5,7,9). 10 is not a digit, its a combination of 2 digits.
4d means that the last digit has to be odd.

so the answer would be 1*4*3 = 12? is it right?

 

[Erzeon]

for 4c or 4d? , for 4c the answer is 120. it should be anyway

 

[six789]

for 4c or 4d? , for 4c the answer is 120. it should be anyway

how could that be? it is 5!/1!?

 

[Erzeon]

(5 x 4 x 3 x 2 x 1) / 1 = 5!/1!

 

[six789]

so I am right, lol

 

[six789]

how about for no. 3a, to see if i get it, ill try to answer it and then check it if is right..

 

[six789]

10!/(2!x2!x2!x1!x1!x1!x1!)
or just 10!/((2!x2!x2!)

 

[Erzeon]

which question is no.7?

 

[six789]

3a.....

 

[six789]

lol..........

 

[Erzeon]

yes 10!/(2! x 2! x 2!) should be correct

 

[six789]

claps, I am right... its really amazing when you get to know how to solve a problem!

 

[six789]

for 3b, is it 7!/(2!x2!x2!) and for 3c is it 6!/((2!x2!x2!))?

 

[Erzeon]

3b should be 9!/(2! x 2! x 2!) there are 10 letters and K must be first
3c should be 9!/(2! x 2!)

 

[six789]

damn, i wrong here... but i get the answer, i just need some more practice and ill be good at it...

 

[six789]

how about for 4b, it is 7x6x5? and for 4d&e is pretty same thiing as 4c?

 

[Erzeon]

4b is 7x7x6x5 because you can choose 7(1 to 7, not including 0) numbers, then 7(you can choose from 0-7 but not the number you chose for your first digit) numbers then 6 numbers then 5 number to produce your 4 digits

 

[Erzeon]

and no, 4c and d and e are not the same. 4d and e might be though.

 

[six789]

so for 4d, is it 5!/(4!)? since there is only 5 numbers (ie 1,3,5,7,9) and there is only 4 places you can switch it...

 

[Erzeon]

4d is saying that you can use digits from 0-9 but the last digit must be odd.

 

[six789]

4d is saying that you can use digits from 0-9 but the last digit must be odd.

is that so... so can you repeat the same number? and why should the number (odd or even) be the last?

 

[Erzeon]

From the question you wrote, you can't repeat the same number, the last number should be odd because when it's odd, it produces an odd number.

 

[six789]

From the question you wrote, you can't repeat the same number,

ok, i get it...

but this one i don't get it...the last number should be odd because when it's odd, it produces an odd number

 

[Erzeon]

i mean last digit, odd numbers always have their last digit as an odd number.

 

[six789]

ohh, i know what you mean now... so the answer in 4d would be...still figuring out...lol

 

[six789]

is it 9!/5! or 9x8x7x6 for 4d? since the numbers runs from 0 to 9, but there is only 5 odd numbers, and same thing with 4e?